(1)Mendel crossed pure-breeding pea plants grown from round seeds with purple flowers with pure-breeding plants grown from wrinkled seeds with white flowers. All the progeny were round seed, purple flowers. Explain this with a suitable genetic diagram.
I drew a punett square and got all to be round seed, purple flower.)
2)When the plants from Q1 were self-pollinated the following season, the following seeds were produced
Phenotype |Genotype|N° of seeds
round,purple | ? |317
round,white | ? |105
wrinkled,purple| ? |106
wrinkled,white | ? |36
a)Which characteristics are not seen in the parents?What does this mean?
b)What is the ratio of phenotypes produced?
c)Draw a genetic diagram justifying these results
d)Cross the heterozygous dominant F1 with the homozygous recessive pea plant. What ratio and phenotypes are produced?Can you think of a use for this?
PLEASE HELP!!!explanations wud be gud
a) The characteristics not seen in the parents are the round, white phenotype and the wrinkled, purple phenotype. This means that these traits are recessive and were masked by the dominant traits in the parental plants.
b) To determine the ratio of phenotypes produced, we can use the observed number of each phenotype and divide it by the total number of seeds produced.
- Round, purple: 317 seeds
- Round, white: 105 seeds
- Wrinkled, purple: 106 seeds
- Wrinkled, white: 36 seeds
To find the ratio, we can simplify the numbers by dividing them all by the smallest number (which is 36 in this case) to get:
- Round, purple: 317/36 ≈ 8.8
- Round, white: 105/36 ≈ 2.9
- Wrinkled, purple: 106/36 ≈ 2.9
- Wrinkled, white: 36/36 = 1
Therefore, the ratio of phenotypes produced is approximately 8.8:2.9:2.9:1, which can be simplified to 3:1:1:1.
c) To draw a genetic diagram explaining these results, we can use Punnett squares. We can represent the round seed and purple flower as dominant traits (R and P) and the wrinkled seed and white flower as recessive traits (r and p):
RrPp x RrPp
--------
RP | RP RP
rP | rP rP
Rp | Rp Rp
rp | rp rp
In this F2 generation cross, we can see that the resulting phenotypes are in a 3:1:1:1 ratio, matching the phenotypes observed in the question.
d) Crossing the heterozygous dominant F1 (RrPp) with the homozygous recessive pea plant (rrpp) can yield the following phenotypes:
RrPp x rrpp
--------
RP | RrPp RrPp
rP | rrp rrp
Rp | RrPp RrPp
rp | rrp rrp
The resulting phenotypes will be:
- Round, purple: RrPp, which appears twice
- Round, white: rrp, which also appears twice
Therefore, the ratio of phenotypes produced will be 2:2 or 1:1, with equal numbers of round, purple and round, white offspring.
One potential use for this breeding is to observe and study the inheritance of certain traits. By crossing a known heterozygous dominant individual with a homozygous recessive individual, we can better understand the patterns of inheritance and determine if the dominant phenotype is due to one or more genes.